A freshly prepared smaple contains `16xx10^(20)` raadioactive nuclei, whose mean life is `10^(10)` seconds The acitiivity of the sample just after 4 ahlr lives time is

Options

A. `(ln2)/(16)xx10^(10)dps`
B. `1xx10^(10) dps`
C. `(1)/(ln2)xx10^(10) dps`
D. `ln 2xx10^(10) dps`

Answer

To solve the problem, we need to find the activity of a radioactive sample just after 4 half-lives. Here's the step-by-step solution:

### Step 1: Understand the given data
- Initial number of radioactive nuclei, \( N_0 = 16 \times 10^{20} \)
- Mean life (average lifetime), \( \tau = 10^{10} \) seconds

### Step 2: Calculate the half-life
The half-life \( T_{1/2} \) is related to the mean life \( \tau \) by the formula:
\[ T_{1/2} = \tau \ln(2) \]
However, for this problem, we can directly use the relationship that after each half-life, the number of undecayed nuclei is halved.

### Step 3: Calculate the number of undecayed nuclei after 4 half-lives
After each half-life, the number of undecayed nuclei is halved:
- After 1 half-life: \( N_0/2 \)
- After 2 half-lives: \( N_0/4 \)
- After 3 half-lives: \( N_0/8 \)
- After 4 half-lives: \( N_0/16 \)

Thus, the number of undecayed nuclei after 4 half-lives is:
\[ N = \frac{N_0}{16} = \frac{16 \times 10^{20}}{16} = 10^{20} \]

### Step 4: Calculate the decay constant \( \lambda \)
The decay constant \( \lambda \) is given by:
\[ \lambda = \frac{1}{\tau} \]
Substituting the value of \( \tau \):
\[ \lambda = \frac{1}{10^{10}} \text{ seconds}^{-1} \]

### Step 5: Calculate the activity of the sample
The activity \( A \) of a radioactive sample is given by:
\[ A = \lambda N \]
Substituting the values of \( \lambda \) and \( N \):
\[ A = \left(\frac{1}{10^{10}}\right) \times (10^{20}) = 10^{10} \text{ disintegrations per second (dps)} \]

### Final Answer
The activity of the sample just after 4 half-lives is:
\[ A = 1 \times 10^{10} \text{ dps} \]